Question: Find the only real number that can be expressed in the form \[(a + bi)^3 - 107i,\]where $i^2 = -1,$ and $a$ and $b$ are positive integers.
Explanation: We have \[\begin{aligned} (a+bi)^3 - 107i &= (a^3 + 3a^2bi - 3ab^2 - b^3i) - 107i \\ &=(a^3 - 3ab^2) + (3a^2b-b^3-107)i. \end{aligned}\]If this is a real number, then we must have \[0 = 3a^2b-b^3-107\]or \[107 = b(3a^2-b^2).\]Since $107$ is prime, either $b=1$ or $b=107.$ If $b=1,$ then we have $107 = 3a^2-1,$ so $a^2 = 36$ and $a=6.$ If $b = 107,$ then we have $1 = 3a^2 - 107^2,$ so $a^2 = \frac{1 + 107^2}{3}.$ But $107^2 \equiv 2^2 \equiv 1 \pmod{3},$ so the right-hand side is not an integer. Thus, $(a, b) = (6, 1)$ is the only possibility. Then the answer is \[a^3 - 3ab^2 = 6^3-3 \cdot 6 \cdot 1^2 = \boxed{198}.\]